Question

An particle moves in a circular path of radius 0.83 cm in the presence of a magnetic field of 0.25 Wb $m^2$. The de Broglie wavelength associated with the particle will be :

• A. $1\stackrel{0}{A}$
• B. $0.1\stackrel{0}{A}$
• C. $10\stackrel{0}{A}$
• D. $0.01\stackrel{0}{A}$

Answer 1

The correct option is D $0.01\stackrel{0}{A}$

Radius of the circular path of a changed particle in a magnetic field is given by

$R=\frac{mv}{Bq}ormv=RBq$

Here, $R=0.83cm=0.83\times {10}^{2}m$

$B=0.25Wb{m}^{-2}$

$q=2e=2\times 1.6\times {10}^{-19}C$

$\therefore mv=(0.83\times {10}^{-2})\left(0.25\right)(2\times 1.6\times {10}^{-19})$

de Broglie wavelength,

$\lambda =\frac{h}{mv}=\frac{6.6\times {10}^{-34}}{0.83\times {10}^{-2}\times 0.25\times 2\times 1.6\times {10}^{-19}}=0.01A$